Positive solutions for higher order differential equations with integral boundary conditions

Abstract

In this paper, we consider the existence of at least three positive solutions for the 2nth order differential equations with integral boundary conditions

$$ \left \{ \textstyle\begin{array}{l} x^{(2n)}(t)=f(t, x(t), x''(t),\ldots,x^{(2(n-1))}(t)), \quad 0\leq t\leq 1, \\ x^{(2i)}(0)=\int_{0}^{1}k_{i}(s)x^{(2i)}(s) \,\mathrm{d}s,\qquad x^{(2i)}(1)=0, \quad 0\leq i\leq n-1, \end{array}\displaystyle \right . $$

where $(-1)^{n}f>0$ is continuous, and $k_{i}(t)\in L^{1}[0,1]$ ($i=0,1,\ldots,n-1$) are nonnegative. The associated Green’s function for the higher order differential equations with integral boundary conditions is first given, and growth conditions are imposed on f which yield the existence of multiple positive solutions by using the Leggett-Williams fixed point theorem.

1 Introduction

The multi-point boundary value problems (BVPs) for ordinary differential equations arise in a variety of different areas of applied mathematics and physics. The study of nonlocal BVPs for second order ordinary differential equations has been widely investigated in [1–3]. Since then, nonlinear high order nonlocal BVPs have been studied by many authors. We refer the reader to [4–13] and references therein. Recently, Guo et al. [14] used Leggett-Williams fixed point theorem to obtain the existence of at least three positive solutions for the 2nth order m-point BVP

$$ \left \{ \textstyle\begin{array}{l} y^{(2n)}(t)=f(t, y(t), y''(t),\ldots,y^{(2(n-1))}(t)), \quad 0\leq t\leq1, \\ y^{(2i)}(0)=0, \qquad y^{(2i)}(1)=\sum_{j=1}^{m-2}k_{ij}y^{(2i)}(\xi_{j}), \quad 0\leq i\leq n-1, \end{array}\displaystyle \right . $$

where $k_{ij}>0$ ($i=0,1,\ldots,n-1$; $j=1,2,\ldots,m-2$), $0<\xi_{1}<\xi_{2}<\cdots<\xi_{m-2}<1$, and $(-1)^{n}f: [0,1]\times\mathbb{R}^{n}\rightarrow[0, +\infty)$ is continuous.

BVPs with integral boundary conditions for ordinary differential equations represent a very interesting and important class of problems and arise in the study of various physical, biological and chemical processes, such as heat conduction, chemical engineering, thermo-elasticity, underground water flow, population dynamics, and plasma physics. Such problems include two-, three-, multi-point and nonlocal BVPs as special cases. The existence and multiplicity of positive solutions for such problems have received a great deal of attention; see [15–19] and references therein. In particular, we would mention the result of [19], Zhang and Ge investigated the existence and nonexistence of positive solutions of the following fourth-order BVP with integral boundary conditions

$$ \left \{ \textstyle\begin{array}{l} x^{(4)}(t)=\omega(t)f(t, x(t), x''(t)), \quad 0< t< 1, \\ x(0)=\int_{0}^{1}g(s)x(s) \,\mathrm{d}s,\qquad x(1)=0, \\ x''(0)=\int_{0}^{1}h(s)x''(s) \,\mathrm{d}s, \qquad x''(1)=0, \end{array}\displaystyle \right . $$

where ω may be singular at $t=0$ and (or) $t=1$, $f\in C([0,1]\times[0,+\infty)\times(-\infty,0],[0,+\infty))$, and $g, h\in L^{1}[0,1]$ are nonnegative.

Motivated by [14, 19], in this paper, we consider the existence of at least three positive solutions for the 2nth order differential equations with integral boundary conditions

$$ \left \{ \textstyle\begin{array}{l} x^{(2n)}(t)=f(t, x(t), x''(t),\ldots,x^{(2(n-1))}(t)),\quad 0\leq t\leq1, \\ x^{(2i)}(0)=\int_{0}^{1}k_{i}(s)x^{(2i)}(s) \,\mathrm{d}s, \qquad x^{(2i)}(1)=0, \quad 0\leq i\leq n-1, \end{array}\displaystyle \right . $$

(1.1)

where $(-1)^{n}f>0$ is continuous, and $k_{i}(t)\in L^{1}[0,1]$ ($i=0,1,\ldots,n-1$) are nonnegative.

For more precise conditions on f, let ${(-1)}^{j}[a,b]=[a,b]$ if j is even and ${(-1)}^{j}[a,b]=[-b,-a]$ if j is odd. Let

$$ \prod_{j=0}^{n-1}[a_{j},b_{j}]=[a_{0},b_{0}] \times\cdots\times [a_{n-1},b_{n-1}]. $$

We shall require that

$$ (-1)^{n}f: [0,1]\times \prod_{j=0}^{n-1}(-1)^{j}[0, +\infty)\rightarrow[0, +\infty). $$

We shall suppose the following conditions are satisfied:

(H₁):

$k_{i}(t)\in L^{1}[0,1]$ are nonnegative, and $K_{i}\in [0,1)$, where

$$ K_{i}= \int_{0}^{1}(1-s)k_{i}(s) \,\mathrm{d}s, \quad 0\leq i\leq n-1; $$

(H₂):

$(-1)^{n}f: [0,1]\times \prod_{j=0}^{n-1}(-1)^{j}[0, +\infty)\rightarrow[0, +\infty)$ is continuous.

2 Preliminary results

Definition 2.1

Let E be a Banach space over $\mathbb{R}$. A nonempty convex closed set $K\subset E$ is said to be a cone provided that

• (i) $au\in K$ for all $u\in K$ and all $a\geq0$;

• (ii) $u,-u\in K$ implies $u=0$.

Definition 2.2

The map α is said to be a nonnegative continuous concave functional on K provided that $\alpha:K\rightarrow[0,\infty)$ is continuous and

$$\alpha\bigl(tx+(1-t)y\bigr)\geq t\alpha(x)+(1-t)\alpha(y) $$

for all $x,y\in K$ and $0\leq t \leq1$. Similarly, we say the map γ is a nonnegative continuous convex functional on K provided that $\gamma:K\rightarrow[0,\infty)$ is continuous and

$$\gamma\bigl(tx+(1-t)y\bigr)\leq t\gamma(x)+(1-t)\gamma(y) $$

for all $x,y\in K$ and $0\leq t \leq1$.

Definition 2.3

Let $0< a< b$ be given and let α be a nonnegative continuous concave functional on K. Define the convex sets $P_{r}$ and $P(\alpha,a,b)$ by

$$ P_{r}=\bigl\{ x\in K|\Vert x\Vert < r\bigr\} \quad \mbox{and} \quad P(\alpha,a,b)=\bigl\{ x\in K|a\leq\alpha(x),\|x\|\leq b\bigr\} . $$

Theorem 2.4

(Leggett-Williams fixed point theorem [20])

Let $A:\overline{P}_{c}\rightarrow\overline{P}_{c}$ be a completely continuous operator and let α be a nonnegative continuous concave functional on K such that $\alpha(x)\leq\|x\|$ for all $x\in\overline{P}_{c}$. Suppose there exist $0< a< b< d\leq c$ such that

(C₁):

$\{x\in P(\alpha,b,d)|\alpha(x)>b\}\neq\emptyset$, and $\alpha(Ax)>b$ for $x\in P(\alpha,b,d)$,

(C₂):

$\|Ax\|< a$ for $\|x\|\leq a$, and

(C₃):

$\alpha(Ax)>b$ for $x\in P(\alpha,b,c)$, with $\|Ax\| >d$.

Then A has at least three fixed points $x_{1}$, $x_{2}$, and $x_{3}$ such that

$$ \|x_{1}\|< a,\qquad b< \alpha(x_{2}) \quad \textit{and} \quad \|x_{3}\|>a \quad \textit{with } \alpha (x_{3})< b . $$

Remark 2.5

If we have $d=c$, then condition (C₁) of Theorem 2.4 implies condition (C₃) of Theorem 2.4.

3 Preliminary lemmas

Lemma 3.1

Suppose (H₁) holds. Then $g_{i}(t,s)\leq0$ ($0\leq i\leq n-1$), where $g_{i}(t,s)$ is the Green’s function for the problem

$$ \left \{ \textstyle\begin{array}{l} x''(t)=0,\quad 0\leq t\leq1, \\ x(0)=\int_{0}^{1}k_{i}(s)x(s) \,\mathrm{d}s, \qquad x(1)=0. \end{array}\displaystyle \right . $$

Proof

It is easy to see that $g_{i}(t,s)\leq0$ ($0\leq i\leq n-1$) by using Lemma 2.1 of [19]. □

Let $G_{1}(t,s)=g_{n-2}(t,s)$, then for $2\leq j\leq n-1$, we recursively define

$$ G_{j}(t,s)= \int_{0}^{1}g_{n-j-1}(t,\tau)G_{j-1}( \tau,s) \,\mathrm{d}\tau. $$

Lemma 3.2

Suppose (H₁) holds. If $y\in C[0,1]$, then the BVP

$$ \left \{ \textstyle\begin{array}{l} x^{(2l)}(t)=y(t),\quad 0\leq t\leq1, \\ x^{(2i)}(0)=\int_{0}^{1}k_{n-l+i-1}(s)x^{(2i)}(s) \,\mathrm{d}s, \qquad x^{(2i)}(1)=0, \quad 0\leq i\leq l-1, \end{array}\displaystyle \right . $$

(3.1)

has a unique solution for each $1\leq l\leq n-1$, where $G_{l}(t, s)$ is the associated Green’s function for the BVP (3.1).

Proof

We will prove the result by using mathematical induction.

When $l=1$, which implies that $i=0$, then the BVP (3.1) reduces to

$$ \left \{ \textstyle\begin{array}{l} x''(t)=y(t),\quad 0\leq t\leq1, \\ x(0)=\int_{0}^{1}k_{n-2}(s)x(s) \,\mathrm{d}s, \qquad x(1)=0. \end{array}\displaystyle \right . $$

(3.2)

By using Lemma 3.1, it is easy to see that the BVP (3.2) has a unique solution

$$ x(t)= \int_{0}^{1}G_{1}(t,s)y(s) \,\mathrm{d}s. $$

Therefore, the result holds for $l=1$.

We assume that the result holds for $l-1$. Now, we deal with the case for l. Let $x''(t)=u(t)$, then the BVP (3.1) is equivalent to the following BVPs:

$$ \left \{ \textstyle\begin{array}{l} x''(t)=u(t), \quad 0\leq t\leq1, \\ x(0)=\int_{0}^{1}k_{n-l-1}(s)x(s) \,\mathrm{d}s,\qquad x(1)=0 \end{array}\displaystyle \right . $$

(3.3)

and

$$ \left \{ \textstyle\begin{array}{l} u^{(2(l-1))}(t)=y(t), \quad 0\leq t\leq1, \\ u^{(2i)}(0)=\int_{0}^{1}k_{n-l+i}(s)u^{(2i)}(s) \,\mathrm{d}s, \qquad u^{(2i)}(1)=0, \quad 0\leq i\leq l-2. \end{array}\displaystyle \right . $$

(3.4)

By applying Lemma 3.1, the BVP (3.3) has a unique solution

$$ x(t)= \int_{0}^{1}g_{n-l-1}(t,r)u(r) \,\mathrm{d}r. $$

(3.5)

Replacing l by $l-1$ and x by u in (3.1), by applying the inductive hypothesis, the BVP (3.4) has also a unique solution

$$ u(t)= \int_{0}^{1}G_{l-1}(t,s)y(s) \,\mathrm{d}s. $$

(3.6)

Substituting (3.6) into (3.5), we see that the BVP (3.1) has a unique solution

$$\begin{aligned} x(t) =& \int_{0}^{1}g_{n-l-1}(t,r) \int_{0}^{1}G_{l-1}(r,s)y(s) \,\mathrm{d}s \,\mathrm{d}r \\ =& \int_{0}^{1} \biggl( \int_{0}^{1}g_{n-l-1}(t,r)G_{l-1}(r,s) \,\mathrm{d}r \biggr)y(s) \,\mathrm{d}s \\ =& \int_{0}^{1}G_{l}(t,s)y(s) \,\mathrm{d}s. \end{aligned}$$

Therefore, the result holds for l. Lemma 3.2 is now completed. □

For each $1\leq l\leq n-1$, we define $A_{l}:C[0,1]\rightarrow C[0,1]$ by

$$ A_{l}u(t)= \int_{0}^{1}G_{l}(t,\tau)u(\tau) \, \mathrm{d}\tau. $$

With the use of Lemma 3.2, for each $1\leq l\leq n-1$, we have

$$ \left \{ \textstyle\begin{array}{l} (A_{l}u)^{(2l)}(t)=u(t),\quad 0\leq t\leq1, \\ (A_{l}u)^{(2i)}(0)=\int _{0}^{1}k_{n-l+i-1}(s)(A_{l}u)^{(2i)}(s) \,\mathrm{d}s, \qquad (A_{l}u)^{(2i)}(1)=0, \quad 0\leq i\leq l-1. \end{array}\displaystyle \right . $$

Therefore (1.1) has a solution if and only if the BVP

$$ \left \{ \textstyle\begin{array}{l} u''(t)=f(t,A_{n-1}u(t),A_{n-2}u(t),\ldots,A_{1}u(t),u(t)), \quad 0\leq t\leq1, \\ u(0)=\int_{0}^{1}k_{n-1}(s)u(s) \,\mathrm{d}s, \qquad u(1)=0 \end{array}\displaystyle \right . $$

(3.7)

has a solution. If x is a solution of (1.1), then $u= x^{(2(n-1))}$ is a solution of (3.7). Conversely, if u is a solution of (3.7), then $x=A_{n-1}u$ is a solution of (1.1). In addition if $(-1)^{n-1}u(t)\geq0$ (≠0) on $[0,1]$, then $x=A_{n-1}u$ is a positive solution of (1.1).

For $0\leq i\leq n-1$, let

$$ m_{i}=\min_{t\in[\tau,1-\tau]} \int_{\tau}^{1-\tau}\bigl\vert g_{i}(t,s)\bigr\vert \,\mathrm{d}s \quad \text{and}\quad M_{i}=\max _{t\in[0,1]} \int_{0}^{1}\bigl\vert g_{i}(t,s)\bigr\vert \,\mathrm{d}s. $$

(3.8)

Obviously, $0< m_{i}<M_{i}$. Let E denote the Banach space $C[0,1]$ with the maximum norm

$$ \|u\|=\max_{0\leq t\leq1}\bigl\vert u(t)\bigr\vert , $$

and define the cone $K \subset E$ by

$$\begin{aligned} K =&\Bigl\{ u\in E\big|(-1)^{n-1}u(t)\geq0, (-1)^{n-1}u(t) \text{ is concave on } [0,1], \text{and} \\ &\min_{t\in[\tau,1-\tau]}(-1)^{n-1}u(t)\geq \tau^{2}\|u \|\Bigr\} . \end{aligned}$$

Finally, we define the nonnegative continuous concave functional α on K by

$$ \alpha(u)=\min_{t\in[\tau,1-\tau]}\bigl\vert u(t)\bigr\vert $$

for each $u\in K$ and it is easy to see that $\alpha(u)\leq\|u\|$.

4 Main results

Theorem 4.1

Suppose conditions (H₁), (H₂) hold. In addition assume there exist nonnegative numbers a, b, and c such that $0< a < b\leq\min \{\tau^{2}, \frac{m_{n-1}}{M_{n-1}} \}c$ and $f(t,u_{n-1},u_{n-2},\ldots,u_{1}, u_{0})$ satisfies the following growth conditions:

(H₃):

$(-1)^{n}f(t,u_{n-1},u_{n-2},\ldots,u_{1}, u_{0})\leq\frac{c}{M_{n-1}}$, for $(t, u_{n-1}, u_{n-2},\ldots, u_{1}, u_{0})\in[0, 1]\times \prod_{j=n-1}^{1}(-1)^{n-1-j} [0, \prod_{i=2}^{j+1}M_{n-i}c ]\times(-1)^{n-1}[0, c]$;

(H₄):

$(-1)^{n}f(t,u_{n-1},u_{n-2},\ldots,u_{1}, u_{0})<\frac{a}{M_{n-1}}$, for $(t, u_{n-1}, u_{n-2},\ldots, u_{1}, u_{0})\in[0, 1]\times \prod_{j=n-1}^{1}(-1)^{n-1-j} [0, \prod_{i=2}^{j+1}M_{n-i}a ]\times(-1)^{n-1}[0, a]$;

(H₅):

$(-1)^{n}f(t,u_{n-1},u_{n-2},\ldots,u_{1}, u_{0})\geq \frac{b}{m_{n-1}}$, for $(t,u_{n-1},u_{n-2},\ldots,u_{1}, u_{0})\in[\tau, 1-\tau]\times \prod_{j=n-1}^{1}(-1)^{n-1-j} [\prod_{i=2}^{j+1}m_{n-i}b, \prod_{i=2}^{j+1}M_{n-i}\frac{b}{\tau^{2}} ]\times (-1)^{n-1}[b, \frac{b}{\tau^{2}}]$.

Then the BVP (1.1) has at least three positive solutions $x_{1}$, $x_{2}$, and $x_{3}$, which satisfy

$$\begin{aligned}& \bigl\Vert x_{1}^{(2(n-1))}(t)\bigr\Vert < a,\qquad b< \alpha \bigl(x_{2}^{(2(n-1))}(t)\bigr)\quad \textit{and} \\& \bigl\Vert x_{3}^{(2(n-1))}(t)\bigr\Vert >a \quad \textit{with } \alpha \bigl(x_{3}^{(2(n-1))}(t)\bigr)< b. \end{aligned}$$

Proof

Define the completely continuous operator A by

$$ Au(t)= \int_{0}^{1}g_{n-1}(t,s)f \bigl(s,A_{n-1}u(s),\ldots,A_{1}u(s),u(s)\bigr) \,\mathrm{d}s. $$

We will first verify that $A:K\rightarrow K$. Let $u\in K$, then $(-1)^{n-1}Au(t)\geq0$, $((-1)^{n-1}Au)''(t)=(-1)^{n-1}f (t,A_{n-1}u(t),\ldots ,A_{1}u(t),u(t) )\leq 0$, $0\leq t\leq1$, and by Proposition 2.2 of [19], we have

$$\begin{aligned} \|Au\| =&\max_{t\in[0,1]}\bigl\vert Au(t)\bigr\vert \\ =&\max_{t\in[0,1]}\biggl\vert \int_{0}^{1}g_{n-1}(t,s)f \bigl(s,A_{n-1}u(s),\ldots,A_{1}u(s),u(s) \bigr) \,\mathrm{d}s \biggr\vert \\ =&\max_{t\in[0,1]}\biggl\vert \int_{0}^{1} \biggl[h(t,s)+\frac {1-t}{1-K_{n-1}} \int_{0}^{1}h(s,\tau)k_{n-1}(\tau) \, \mathrm{d}\tau \biggr] \\ &{}\times f \bigl(s,A_{n-1}u(s),\ldots,A_{1}u(s),u(s) \bigr) \,\mathrm{d}s\biggr\vert \\ \leq& \int_{0}^{1} \biggl[h(s,s)+\frac{1}{1-K_{n-1}} \int_{0}^{1}h(\tau,\tau )k_{n-1}(\tau) \, \mathrm{d}\tau \biggr] \\ &{}\times\bigl|f \bigl(s,A_{n-1}u(s),\ldots,A_{1}u(s),u(s) \bigr)\bigr| \,\mathrm{d}s. \end{aligned}$$

On the other hand, by Proposition 2.3 of [19], we obtain

$$\begin{aligned}& \min_{t\in[\tau,1-\tau]}(-1)^{n-1}Au(t) \\& \quad = \min_{t\in[\tau,1-\tau]}(-1)^{n-1} \int _{0}^{1}g_{n-1}(t,s)f \bigl(s,A_{n-1}u(s),\ldots,A_{1}u(s),u(s) \bigr) \,\mathrm{d}s \\& \quad = \min_{t\in[\tau,1-\tau]}(-1)^{n} \int_{0}^{1} \biggl[h(t,s)+\frac {1-t}{1-K_{n-1}} \int_{0}^{1}h(s,\tau)k_{n-1}(\tau) \, \mathrm{d}\tau \biggr] \\& \qquad {}\times f \bigl(s,A_{n-1}u(s),\ldots,A_{1}u(s),u(s) \bigr) \,\mathrm{d}s \\& \quad \geq \tau^{2} \int_{0}^{1} \biggl[h(s,s)+\frac{1}{1-K_{n-1}} \int_{0}^{1}h(\tau,\tau )k_{n-1}(\tau) \, \mathrm{d}\tau \biggr] \\& \qquad {}\times\bigl\vert f \bigl(s,A_{n-1}u(s), \ldots,A_{1}u(s),u(s) \bigr)\bigr\vert \,\mathrm{d}s \\& \quad \geq \tau^{2}\|Au\|. \end{aligned}$$

Consequently, $A:K\rightarrow K$.

It is a standard argument to show that the operator A is completely continuous. Equicontinuity and uniform boundedness follow readily from the properties of $G_{l}$, $1\leq l\leq n-1$.

If $u\in\overline{P}_{c}$, then $\|u\|\leq c$. For each $1\leq j\leq n-1$, note that inductively (using (3.8)) we have

$$ \|A_{j}u\|=\max_{t\in[0,1]}\biggl\vert \int_{0}^{1}G_{j}(t,s)u(s) \,\mathrm{d}s \biggr\vert \leq\prod_{i=2}^{j+1}M_{n-i}c. $$

From the condition (H₃) and (3.8), we obtain

$$\begin{aligned} \begin{aligned} \|Au\|&=\max_{t\in[0,1]}\bigl\vert Au(t)\bigr\vert \\ &=\max_{t\in[0,1]}\biggl\vert \int_{0}^{1}g_{n-1}(t,s)f \bigl(s,A_{n-1}u(s),A_{n-2}u(s),\ldots,A_{1}u(s),u(s) \bigr) \,\mathrm{d}s\biggr\vert \\ &\leq \frac{c}{M_{n-1}}\max_{t\in[0,1]} \int_{0}^{1}\bigl\vert g_{n-1}(t,s)\bigr\vert \,\mathrm{d}s=c. \end{aligned} \end{aligned}$$

So, $A:\overline{P}_{c}\rightarrow\overline{P}_{c}$.

In a completely analogous argument, the condition (H₄) implies the condition (C₂) of Theorem 2.4 is satisfied.

We now show that condition (C₁) is satisfied. Note that, for $0\leq t\leq1$,

$$ u(t)=(-1)^{n-1}\frac{b}{\tau^{2}}\in P\biggl(\alpha,b,\frac{b}{\tau^{2}} \biggr)\quad \text{and}\quad \alpha(u)=\frac{b}{\tau^{2}}>b. $$

Thus,

$$ \biggl\{ u\in P\biggl(\alpha,b,\frac{b}{\tau^{2}}\biggr)\Big|\alpha(u)>b \biggr\} \neq \emptyset. $$

Also, if $u\in P(\alpha,b,\frac{b}{\tau^{2}})$, then $b\leq (-1)^{n-1}u(t)\leq\frac{b}{\tau^{2}}$ for $t\in[\tau, 1-\tau]$, implies for each $1\leq j\leq n-1$, $t\in[\tau, 1-\tau]$, inductively,

$$\begin{aligned}& (-1)^{n-1-j}A_{j}u(t)\leq\prod_{i=2}^{j+1}M_{n-i} \frac{b}{\tau ^{2}}, \\& (-1)^{n-1-j}A_{j}u(t)=\biggl\vert \int_{0}^{1}G_{j}(t,s)u(s) \,\mathrm{d}s \biggr\vert \geq b \int_{\tau}^{1-\tau}\bigl\vert G_{j}(t,s)\bigr\vert \,\mathrm{d}s\geq\prod_{i=2}^{j+1}m_{n-i}b. \end{aligned}$$

With the use of condition (H₅) and (3.8), we get

$$\begin{aligned} \alpha(Au) =&\min_{t\in[\tau, 1-\tau]} \biggl\vert \int_{0}^{1}g_{n-1}(t,s)f \bigl(s,A_{n-1}u(s),\ldots ,A_{1}u(s),u(s) \bigr) \,\mathrm{d}s \biggr\vert \\ >&\min_{t\in[\tau, 1-\tau]} \biggl\vert \int_{\tau}^{1-\tau}g_{n-1}(t,s)f \bigl(s,A_{n-1}u(s),\ldots ,A_{1}u(s),u(s) \bigr) \,\mathrm{d}s \biggr\vert \\ \geq&\frac{b}{m_{n-1}}\min_{t\in[\tau, 1-\tau]} \int_{0}^{1}\bigl\vert g_{n-1}(t,s)\bigr\vert \,\mathrm{d}s=b. \end{aligned}$$

Therefore, condition (C₁) is satisfied.

Finally, we show that condition (C₃) is also satisfied. That is, we show that if $u\in P(\alpha,b,c)$ and $\|Au\|>d=\frac{b}{\tau^{2}}$, then $\alpha(Au)>b$. This follows since $A:K\rightarrow K$. In particular, since $(-1)^{n-1}(Au)$ is concave and $\min_{t\in[\tau, 1-\tau]}(-1)^{n-1}(Au)(t)\geq \tau^{2}\|Au\|$.

That is,

$$ \alpha(Au)=\min_{t\in[\tau, 1-\tau]}\bigl\vert Au(t)\bigr\vert \geq \tau^{2}\|Au\|>b. $$

Therefore, condition (C₃) is also satisfied. By Theorem 2.4, there exist three solutions $u_{1}, u_{2}, u_{3}\in K$ for the BVP (3.7). Moreover, let

$$ x_{i}(t)=A_{n-1}u_{i}(t)= \int_{0}^{1}G_{n-1}(t,s)u_{i}(s) \,\mathrm{d}s,\quad i=1,2,3, $$

then $x_{1}$, $x_{2}$, $x_{3}$ are three positive solutions for the BVP (1.1) and satisfy

$$\begin{aligned}& \bigl\Vert x_{1}^{(2(n-1))}(t)\bigr\Vert < a,\qquad b< \alpha \bigl(x_{2}^{(2(n-1))}(t)\bigr)\quad \text{and} \\& \bigl\Vert x_{3}^{(2(n-1))}(t)\bigr\Vert >a\quad \text{with } \alpha \bigl(x_{3}^{(2(n-1))}(t)\bigr)< b. \end{aligned}$$

 □

Consider the following 2nth order differential equations with integral boundary conditions:

$$ \left \{ \textstyle\begin{array}{l} x^{(2n)}(t)=f(t, x(t), x''(t),\ldots,x^{(2(n-1))}(t)), \quad 0\leq t\leq 1, \\ x^{(2i)}(0)=0,\qquad x^{(2i)}(1)=\int_{0}^{1}k^{\ast }_{i}(s)x^{(2i)}(s) \,\mathrm{d}s, \quad 0\leq i\leq n-1, \end{array}\displaystyle \right . $$

(4.1)

where $(-1)^{n}f\in C([0,1]\times \prod_{j=0}^{n-1}(-1)^{j}[0, +\infty)\rightarrow[0, +\infty))$ and $k^{\ast}_{i}(t)\in L^{1}[0,1]$ ($i=0,1,\ldots,n-1$) are nonnegative.

Now we deal with problem (4.1). The method is just similar to what we have done for the problem (1.1), so we omit the proof of main results in this section.

For convenience, we list the following assumptions:

($\mathrm{H}^{\ast}_{1}$):

$k^{\ast}_{i}(t)\in L^{1}[0,1]$ are nonnegative, and $K^{*}_{i}\in[0,1)$, where

$$ K^{*}_{i}= \int_{0}^{1}sk^{\ast}_{i}(s) \, \mathrm{d}s, \quad 0\leq i\leq n-1. $$

By analogous methods, we have the following results.

Lemma 4.2

Suppose ($\mathrm{H}^{\ast}_{1}$) holds. Then $g^{\ast }_{i}(t,s)\leq0$ ($0\leq i\leq n-1$), where $g^{\ast}_{i}(t,s)$ is the Green’s function for the problem

$$ \left \{ \textstyle\begin{array}{l} x''(t)=0,\quad 0\leq t\leq1, \\ x(0)=0, \qquad x(1)=\int_{0}^{1}k^{\ast}_{i}(s)x(s) \,\mathrm{d}s. \end{array}\displaystyle \right . $$

For $0\leq i\leq n-1$, let

$$ m^{\ast}_{i}=\min_{t\in[\tau,1-\tau]} \int_{\tau}^{1-\tau}\bigl\vert g^{\ast }_{i}(t,s) \bigr\vert \,\mathrm{d}s \quad \textit{and}\quad M^{\ast}_{i}= \max_{t\in[0,1]} \int_{0}^{1}\bigl\vert g^{\ast}_{i}(t,s) \bigr\vert \,\mathrm{d}s. $$

Theorem 4.3

Suppose condition ($\mathrm{H}^{\ast}_{1}$) holds. In addition assume there exist nonnegative numbers a, b, and c such that $0< a < b\leq\min \{\tau^{2}, \frac{m^{\ast}_{n-1}}{M^{\ast}_{n-1}} \}c$ and $f(t,u_{n-1},u_{n-2},\ldots,u_{1}, u_{0})$ satisfies the following growth conditions:

($\mathrm{H}^{\ast}_{2}$):

$(-1)^{n}f(t,u_{n-1},u_{n-2},\ldots,u_{1}, u_{0})\leq\frac{c}{M^{\ast}_{n-1}}$, for $(t, u_{n-1}, u_{n-2},\ldots, u_{1}, u_{0})\in[0, 1]\times \prod_{j=n-1}^{1}(-1)^{n-1-j} [0, \prod_{i=2}^{j+1}M^{\ast}_{n-i}c ]\times(-1)^{n-1}[0, c]$;

($\mathrm{H}^{\ast}_{3}$):

$(-1)^{n}f(t,u_{n-1},u_{n-2},\ldots,u_{1}, u_{0})<\frac{a}{M^{\ast}_{n-1}}$, for $(t, u_{n-1}, u_{n-2},\ldots, u_{1}, u_{0})\in[0, 1]\times \prod_{j=n-1}^{1}(-1)^{n-1-j} [0, \prod_{i=2}^{j+1}M^{\ast}_{n-i}a ]\times(-1)^{n-1}[0, a]$;

($\mathrm{H}^{\ast}_{4}$):

$(-1)^{n}f(t,u_{n-1},u_{n-2},\ldots,u_{1}, u_{0})\geq\frac{b}{m^{\ast}_{n-1}}$, for $(t,u_{n-1},u_{n-2},\ldots,u_{1}, u_{0})\in[\tau, 1-\tau]\times \prod_{j=n-1}^{1}(-1)^{n-1-j} [\prod_{i=2}^{j+1}m^{\ast}_{n-i}b, \prod_{i=2}^{j+1}M^{\ast}_{n-i}\frac{b}{\tau^{2}} ]\times (-1)^{n-1}[b, \frac{b}{\tau^{2}}]$.

Then the BVP (4.1) has at least three positive solutions $x_{1}$, $x_{2}$, and $x_{3}$, which satisfy

$$\begin{aligned}& \bigl\Vert x_{1}^{(2(n-1))}(t)\bigr\Vert < a,\qquad b< \alpha \bigl(x_{2}^{(2(n-1))}(t)\bigr)\quad \textit{and} \\& \bigl\Vert x_{3}^{(2(n-1))}(t)\bigr\Vert >a \quad \textit{with } \alpha \bigl(x_{3}^{(2(n-1))}(t)\bigr)< b. \end{aligned}$$

5 Example

Example 5.1

As an example of problem (1.1), consider the following sixth order BVP:

$$ \left \{ \textstyle\begin{array}{l} x^{(6)}(t)=f(t,x(t),x''(t),x^{(4)}(t)), \quad 0\leq t\leq1, \\ x(0)=\int^{1}_{0}x(s) \,\mathrm{d}s, \qquad x(1)=0, \\ x''(0)=\int^{1}_{0}sx''(s) \,\mathrm{d}s,\qquad x''(1)=0, \\ x^{(4)}(0)=\int^{1}_{0}s^{2}x^{(4)}(s) \,\mathrm{d}s,\qquad x^{(4)}(1)=0, \end{array}\displaystyle \right . $$

(5.1)

where

$$\begin{aligned}& f(t,u_{2},u_{1},u_{0}) \\& \quad = \left \{ \textstyle\begin{array}{l@{\quad}l} -\frac{6\text{,}050}{841}u_{0}\cdot\frac{2+\sin (t+u_{1}+u_{2})}{3},&0\leq u_{0}\leq1, \\ - [ (6\times\frac{112\text{,}640}{7\text{,}559}-\frac {6\text{,}050}{841} )(u_{0}-1)+\frac{6\text{,}050}{841} ]\cdot\frac{2+\sin (t+u_{1}+u_{2})}{3}, &1\leq u_{0}\leq2, \\ -3\times\frac{112\text{,}640}{7\text{,}559}u_{0}\cdot \frac{2+\sin(t+u_{1}+u_{2})}{3}, &2\leq u_{0}\leq32, \\ - [96\times\frac{112\text{,}640}{7\text{,}559}+ (\frac{8}{7}\times\frac {6\text{,}050}{841}-\frac{3}{7}\times\frac{112\text{,}640}{7\text{,}559} )(u_{0}-32) ] \\ \quad {}\cdot\frac{2+\sin(t+u_{1}+u_{2})}{3},&32\leq u_{0}\leq256, \\ -256\times\frac{6\text{,}050}{841}\cdot \frac{2+\sin(t+u_{1}+u_{2})}{3}, &u_{0}\geq256. \end{array}\displaystyle \right . \end{aligned}$$

We notice that $n=3$, $k_{i}(s)=s^{i}$ ($i=0, 1, 2$) and $K_{0}=\frac{1}{2}$, $K_{1}=\frac{1}{6}$, $K_{2}=\frac{1}{12}$.

If we take $\tau=\frac{1}{4}$, by calculation we obtain

$$\begin{aligned}& m_{0}=\min_{t\in[\frac{1}{4},\frac{3}{4}]} \int_{\frac {1}{4}}^{\frac{3}{4}}\bigl\vert g_{0}(t,s)\bigr\vert \,\mathrm{d}s=\frac{35}{384}, \qquad M_{0}=\max _{t\in[0,1]} \int _{0}^{1}\bigl\vert g_{0}(t,s)\bigr\vert \,\mathrm{d}s=\frac{2}{9}, \\& m_{1}=\min_{t\in[\frac{1}{4},\frac{3}{4}]} \int_{\frac {1}{4}}^{\frac{3}{4}}\bigl\vert g_{1}(t,s)\bigr\vert \,\mathrm{d}s=\frac{97}{1\text{,}280},\qquad M_{1}=\max _{t\in[0,1]} \int_{0}^{1}\bigl\vert g_{1}(t,s)\bigr\vert \,\mathrm{d}s=\frac{121}{800}, \\& m_{2}=\min_{t\in[\frac{1}{4},\frac{3}{4}]} \int_{\frac {1}{4}}^{\frac{3}{4}}\bigl\vert g_{2}(t,s)\bigr\vert \,\mathrm{d}s=\frac{7\text{,}559}{112\text{,}640},\qquad M_{2}=\max _{t\in[0,1]} \int_{0}^{1}\bigl\vert g_{2}(t,s)\bigr\vert \,\mathrm{d}s=\frac{841}{6\text{,}050}. \end{aligned}$$

In addition, if we take $a=1$, $b=2$, $c=256$, then

$$\begin{aligned} 0 < &a=1< b=2\leq\min \biggl\{ \tau^{2}, \frac{m_{2}}{M_{2}} \biggr\} c \\ =&\min \biggl\{ \biggl(\frac{1}{4} \biggr)^{2}, \frac{7\text{,}559\times6\text{,}050}{112\text{,}640\times841} \biggr\} \times256=16, \end{aligned}$$

and $f(t,u_{2},u_{1},u_{0})$ satisfies the growth conditions (H₃)-(H₅).

Therefore all the conditions of Theorem 4.1 are satisfied. Hence, the problem (5.1) has at least three positive solutions $x_{1}$, $x_{2}$, and $x_{3}$, which satisfy

$$\begin{aligned} \begin{aligned} &\max_{0\leq t\leq 1}\bigl\vert x_{1}^{(4)}(t)\bigr\vert < 1,\qquad 2< \min_{t\in[\frac{1}{4},\frac {3}{4}]}\bigl\vert x_{2}^{(4)}(t) \bigr\vert \quad \text{and} \\ &\max_{0\leq t\leq1}\bigl\vert x_{3}^{(4)}(t)\bigr\vert >1\quad \text{with } \min _{t\in[\frac{1}{4},\frac{3}{4}]}\bigl\vert x_{3}^{(4)}(t)\bigr\vert < 2. \end{aligned} \end{aligned}$$

Example 5.2

As another example of problem (4.1), consider the following sixth order BVP:

$$ \left \{ \textstyle\begin{array}{l} x^{(6)}(t)=f(t,x(t),x''(t),x^{(4)}(t)), \quad 0\leq t\leq1, \\ x(0)=0, \qquad x(1)=\int^{1}_{0}x(s) \,\mathrm{d}s, \\ x''(0)=0, \qquad x''(1)=\int^{1}_{0}sx''(s) \,\mathrm{d}s, \\ x^{(4)}(0)=0, \qquad x^{(4)}(1)=\int ^{1}_{0}s^{2}x^{(4)}(s) \,\mathrm{d}s, \end{array}\displaystyle \right . $$

(5.2)

where

$$\begin{aligned}& f(t,u_{2},u_{1},u_{0}) \\& \quad = \left \{ \textstyle\begin{array}{l@{\quad}l} -\frac{225}{32}u_{0}\cdot\frac{2+\sin (t+u_{1}+u_{2})}{3},& 0\leq u_{0}\leq1, \\ - [ (6\times\frac{30\text{,}720}{2\text{,}097}-\frac {225}{32} )(u_{0}-1)+\frac{225}{32} ]\cdot\frac{2+\sin (t+u_{1}+u_{2})}{3}, &1\leq u_{0}\leq2, \\ -3\times\frac{30\text{,}720}{2\text{,}097}u_{0}\cdot \frac{2+\sin(t+u_{1}+u_{2})}{3}, & 2\leq u_{0}\leq32, \\ - [96\times\frac{30\text{,}720}{2\text{,}097}+ (\frac{8}{7}\times\frac {225}{32}-\frac{3}{7}\times\frac{30\text{,}720}{2\text{,}097} )(u_{0}-32) ] \cdot\frac{2+\sin(t+u_{1}+u_{2})}{3},& 32\leq u_{0}\leq256, \\ -256\times\frac{225}{32}\cdot \frac{2+\sin(t+u_{1}+u_{2})}{3}, & u_{0}\geq256. \end{array}\displaystyle \right . \end{aligned}$$

We notice that $n=3$, $k^{\ast}_{i}(s)=s^{i}$ ($i=0, 1, 2$) and $K^{\ast}_{0}=\frac{1}{2}$, $K^{\ast}_{1}=\frac{1}{3}$, $K^{\ast}_{2}=\frac{1}{4}$.

If we take $\tau=\frac{1}{4}$, by calculation we obtain

$$\begin{aligned}& m^{\ast}_{0}=\min_{t\in[\frac{1}{4},\frac{3}{4}]} \int _{\frac{1}{4}}^{\frac{3}{4}}\bigl\vert g^{\ast}_{0}(t,s) \bigr\vert \,\mathrm{d}s=\frac{35}{384},\qquad M^{\ast}_{0}= \max_{t\in[0,1]} \int _{0}^{1}\bigl\vert g^{\ast}_{0}(t,s) \bigr\vert \,\mathrm{d}s=\frac{2}{9}, \\& m^{\ast}_{1}=\min_{t\in[\frac{1}{4},\frac{3}{4}]} \int _{\frac{1}{4}}^{\frac{3}{4}}\bigl\vert g^{\ast}_{1}(t,s) \bigr\vert \,\mathrm{d}s=\frac{123}{1\text{,}024},\qquad M^{\ast}_{1}= \max_{t\in[0,1]} \int_{0}^{1}\bigl\vert g^{\ast}_{1}(t,s) \bigr\vert \,\mathrm{d}s=\frac{81}{512}, \\& m^{\ast}_{2}=\min_{t\in[\frac{1}{4},\frac{3}{4}]} \int _{\frac{1}{4}}^{\frac{3}{4}}\bigl\vert g^{\ast}_{2}(t,s) \bigr\vert \,\mathrm{d}s=\frac{2\text{,}097}{30\text{,}720},\qquad M^{\ast}_{2}= \max_{t\in[0,1]} \int_{0}^{1}\bigl\vert g^{\ast}_{2}(t,s) \bigr\vert \,\mathrm{d}s=\frac{32}{225}. \end{aligned}$$

In addition, if we take $a=1$, $b=2$, $c=256$, then

$$ 0< a=1< b=2\leq\min \biggl\{ \tau^{2}, \frac{m^{\ast}_{2}}{M^{\ast}_{2}} \biggr\} c=\min \biggl\{ \biggl(\frac {1}{4} \biggr)^{2}, \frac{2\text{,}097\times225}{30\text{,}720\times32} \biggr\} \times256=16, $$

and $f(t,u_{2},u_{1},u_{0})$ satisfies the growth conditions ($\mathrm{H}_{2}^{\ast}$)-($\mathrm{H}_{4}^{\ast}$).

Therefore all the conditions of Theorem 4.3 are satisfied. Hence, the problem (5.2) has at least three positive solutions $x_{1}$, $x_{2}$, and $x_{3}$, which satisfy

$$\begin{aligned} \begin{aligned} &\max_{0\leq t\leq 1}\bigl\vert x_{1}^{(4)}(t)\bigr\vert < 1,\qquad 2< \min_{t\in[\frac{1}{4},\frac {3}{4}]}\bigl\vert x_{2}^{(4)}(t) \bigr\vert \quad \text{and} \\ &\max_{0\leq t\leq1}\bigl\vert x_{3}^{(4)}(t)\bigr\vert >1\quad \text{with } \min _{t\in[\frac{1}{4},\frac{3}{4}]}\bigl\vert x_{3}^{(4)}(t)\bigr\vert < 2. \end{aligned} \end{aligned}$$